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Alkene Reactions

By James Ashenhurst

A Fourth Alkene Addition Pattern – Free Radical Addition

Last updated: July 9th, 2019 |

Free Radical Addition Of HBr To Alkenes With ROOR (Peroxides) 

We’ve seen that there are three major alkene reactivity patterns [carbocation, three membered ring, and concerted], but there are two minor pathways as well. This post discusses one of them: free-radical addition of HBr to alkenes, which shows the opposite regioselectivity (anti-Markovnikov) than “normal” addition of HBr to alkenes (Markovnikov) which follows the “carbocation” pathway.

Table of Contents

  1. Free Radical Addition Of HBr To Alkenes Leads To “Anti-Markovnikov” Products
  2. An Outline Of The Free Radical Mechanism For Addition Of HBr To Alkenes In The Presence Of Peroxides
  3. Initiation Of The Free-Radical Process Through Homolytic Cleavage Of Peroxides By Heat Or Light
  4. Formation Of The Bromine Radical From The Alkoxide Radical And  HBr
  5. Propagation Step #1 : Addition Of Bromine Radical To The Alkene Occurs So As To Give The Most Stable Carbon Radical
  6. Propagation Step #2: The Resulting Carbon Radical Removes A Hydrogen Atom From H–Br, Regenerating The Bromine Radical
  7. The Termination Step
  8. Summary: Free-Radical Addition Of HBr To Alkenes
  9. Notes

1. Free Radical Addition Of HBr To Alkenes Leads To “Anti-Markovnikov” Products

As discussed previously, alkenes normally react with HBr to give products of “Markovnikov” addition; the bromine ends up on the most substituted carbon of the alkene, and the hydrogen ends up on the least substituted carbon. However, something interesting happens when the same reaction is performed in the presence of peroxides and  heat / light: the pattern of addition changes!

Instead of Br ending up on the most substituted carbon of the alkene, it ends up on the least. [The stereochemistry of the reaction, however, is unchanged: it still gives a mixture of “syn” and “anti” products.]

1-alkene hbr

This so-called “anti-Markovnikov” addition is intriguing. What difference could the presence of peroxides, and furthermore heat (or light) make to this reaction?

2. An Outline Of The Free Radical Mechanism For Addition Of HBr To Alkenes In The Presence Of ROOR (Peroxides)

This reaction occurs through a free-radical process. (For a primer on free radical chemistry, you might want to check out this chapter).  Here is an outline of the mechanism:

  • Peroxides contain a weak oxygen-oxygen bond [approximately 35 kcal/mol;  compare to C-H at approx 100 kcal/mol]
  • Heating leads to homolytic fragmentation of this bond – that is, the bond breaks such as to leave one unpaired electron on each atom. Strong sources of light [e.g. a floodlight or other source of light radiation which reaches into the near UV] can also serve to sever this bond.
  • The resulting highly reactive alkoxy radical can then abstract a hydrogen from H-Br, giving a bromine radical. The bromine radical is the species that adds to the alkene.
  • Addition to the alkene will preferably occur in such a way that the most stable free radical is formed [in the case above, the tertiary radical]. That’s why bromine ends up on the least substituted carbon of the alkene. (See: 3 Factors Which Stabilize Free Radicals)
  • This tertiary radical then removes hydrogen from H-Br, liberating a bromine radical, and the cycle continues.

3. Initiation Of The Free-Radical Process Through Homolytic Cleavage Of ROOR (Peroxides) By Heat Or Light

Only a trace [catalytic] amount of peroxide is required to get the reaction started, although of course at least one molar equivalent of HBr is required to result in full addition of HBr to the alkene.

In the first step, addition of energy (in the form of heat or light) leads to homolytic fragmentation of the weak O–O bond to generate two new free radicals.  “Homolytic” means that the bond is broken such that each atom receives the same (“homos” = Greek for “same”) number of electrons.

(Most of the bond breakage we see in organic chemistry is heterolytic, where the bond breaks unequally. )

2-alkene hbr

The singly barbed arrows depict the movement of single electrons; two alkoxy radicals are formed. Since there is a net increase in the number of radicals (0 →2) this is an initiation step.

Common “peroxides” for this purpose are t-butyl peroxide or benzoyl peroxide. * [Note 1]. Alternatively other free-radical “initiators” such as AIBN can also be used.

Only a catalytic amount of peroxides are used to initate this reaction (typically 10-20 mole %, although more can be used, especially when added batchwise).

4. Formation Of The Bromine Radical From The Alkoxide Radical And  HBr

In the next step, one of the oxygen radicals from step 1 removes a hydrogen from H–Br in another homolytic process.

3-hbr

Here, we’re forming an H–O bond (bond dissociation energy of 102 kcal/mol for H–O in CH3OH) and breaking an H–Br bond (bond dissociation energy of 87 kcal/mol) , so a difference in energy of about 15 kcal/mol makes this process essentially irreversible.

(Note: since this process does not change the number of free radicals, it is technically a propagation step)

5. Propagation Step #1 : Addition Of Bromine Radical To The Alkene Occurs So As To Give The Most Stable Carbon Radical

Once formed, the bromine radical can then add to the alkene.

In a relatively “flat” alkene such as 1-methylcyclohexene, addition of the radical will occur with equal probability from either face.

The question is, which atom of the double bond does the free radical attack? The bond could break two different ways, after all.

  • Attack of the bromine radical on the more substituted carbon would result in a new free radical on a secondary carbon.
  • Attack of the bromine radical on the less substituted carbon  would result in a new free radical on a tertiary carbon.

Free radicals are electron-deficient species and are stabilized by adjacent electron donors. The more stable free radical intermediate is the tertiary free radical, and that is why addition occurs predominantly at the less substituted carbon (i.e. the carbon attached to the fewest number of carbons).

This explains the “anti-Markovnikov” selectivity of the reaction.

4-hbr

6. Propagation Step #2: The Resulting Carbon Radical Removes A Hydrogen Atom From H–Br, Regenerating The Bromine Radical

In a second propagation step in the main sequence, the resulting carbon radical removes a hydrogen from another equivalent of H–Br, giving the final addition product.

Alkyl free radicals are sp2-hybridized, and are shallow pyramids that invert easily.

H–Br, therefore, can react on either face of the free radical [note 2]. If it attacks on the same face as the Br, then we obtain a “syn” product. If it attacks on the opposite face of the Br, then the product is “anti“.

A mixture of both will be obtained. The reaction is not stereoselective.

5-hbr

A bromine radical is generated by this process, which can then add to another equivalent of alkene (propagation step #1).

7. The Termination Step

When the concentration of HBr and alkene become low relative to the concentration of free radical, termination can occur. This could occur through a variety of specific pathways (not shown)  involving recombination of two free radicals to generate a new bond.

8. Summary: Free-Radical Addition Of HBr To Alkenes

This reaction pathway is most commonly observed (in Org 1 and Org 2, anyway) for addition of HBr, although a rich chemistry of radical addition reactions to alkenes exists (particularly for organostannanes).

NEXT POST: Ozonolysis of Alkenes


Notes

* Note. Benzoyl peroxide enjoys a common household use as an acne cleanser, and even makes an appearance in this classic ad.

**Note 2. The geometry of free radical carbons is that of a  shallow pyramid with a low barrier for inversion, allowing for reactivity on either face. The exception is in weird cases where inversion would be highly disfavored, such as on a bridgehead.

Further Reading

  1. Bond Dissociation Energies From Lowry & Richardson, “Mechanism and Theory In Organic Chemistry“, Harper & Row, 1987 pp 161-162.

Comments

Comment section

29 thoughts on “A Fourth Alkene Addition Pattern – Free Radical Addition

  1. Penultimate paragraph: “This reaction pathway is most commonly observed (in Org 1 and Org 2, anyway) for” – unfinished sentence.

    Great job, as usual.

    1. Not sure what you mean – you mean react with bromine radical?

      The concentration of bromine radical at any given moment is much less than the concentration of HBr.

      If it reacted with bromine radical that would be a termination step!

      1. No, in fact he is asking about the ‘Abstraction of hydrogen in H-Br…’ step, why don’t we get a dibromo cyclo-product. Well, that reaction would simply be thermodynamically unfavorable (+39 kcal in contrast with -23 kcal). This is discussed on page 387 in the book ‘Basic principles of organic chemistry’ (just Google it, it’s fascinating!). And I hope this helps.

  2. In step 2, why does the bromine radical not abstract an allylic hydrogen (like in NBS halogenation reactions)? I presume that allylic radical would be almost as stable as a tertiary radical.

    1. I think it’s because the C-C pi bond strength is about 60-65 kcal/mol whereas C-H allylic bond strength is about 80-85 kcal/mol. Higher activation energy in other words

      1. Could it be that allylic hydrogen abstraction does occur, but the resulting allyl radical has nothing to react with except H-Br? The allyl radical could reabstract a hydrogen from H-Br, to reform the alkene, which then eventually adds a bromine radical. Potentially the alkene could isomerize if this occurred, and I don’t know if this is observed. Just a thought…

    1. HCL will not form chlorine radical as chlorine is highly electronegative…it will always form a cation only.

  3. In the last step, why the tertiary radical react with H from H-Br not Br from H-Br? (and the product will be 1,2-dibromo-2-methylcyclohexane)

    1. Reaction with Br will give the hydrogen radical, H• , which is considerably less stable than the bromine radical. And formation of the C-Br bond is much weaker than formation of the C-H bond.

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